# Horsepower requirement of the tugboat towing the pipeline

Discussion in 'Boat Design' started by Ali Golshani, Nov 7, 2017.

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### baeckmoHydrodynamics

Hello Ali, you asked for an example, here we go. I'll show two "basic engineering" methods as opposed to our button-pressing phantoms.......;

The resistance comes from two parts, the surface friction and the dynamic stagnation pressure loss. Since you are dealing with pipes, we can use the pipe friction calculation for this example; somewhere, somebody has already done a friction loss estimate (and you could use their numbers directly). It is based on the wetted surface, so imagine your pipe is having an inner dia of 2.75 m. The water does not know whether it is inside or outside the wall. Let’s assume it is floating half submerged.

• First, calculate Reynolds number: Re = V*D/ v; where V=speed, m/s, D=diameter, m, v=kinematic viscosity. Here: Re = 2.75*2/1e-6; dvs Re = 8250000.

• Estimate surface roughness, ks. Say ks ~0.1 mm. Then D/ks = 27500.

• In pipe friction diagram (lambda over Re, according to Nikuradse f.i.) find lambda for your Re and D/ks. In this case, lambda ~0.09.

• Calculate pipe friction loss from pf = lambda*L/D*pdyn; where L = pipe length and pdyn is dynamic pressure, i.e. V^2*density/2. Here pdyn = 4500 Pa. The pressure loss for the fully wetted pipe is thus pf = 2121 Pa. This is the pressure differential needed to move water at 3 m/s (10.8 km/h) through the pipe, against the wall friction. The total force of this pressure over the pipe opening is: Ff = pf * Area. For a half full pipe, the force is ~2121*5.94/2, i.e. Ff =6300 N.

• Next, calculate stagnation pressure loss as ps = pf*Cr; where Cr = loss coefficient, here ~1. The force to overcome is then ps*submerged area. Here Fs = 13365 N.

• The total resistance is then ~20000 N.
Now the friends of marine order will raise their voices and protest, this is a calculation of pipe flow resistance, it can’t be relied upon! So, let’s see where the differences hide then. For hull friction we use the Schoenherr method, which is based on Reynolds number using total length, and total friction area.

• Re = 3*144/1e-6, dvs Re = 4.3e8.

• Roughness correction for L/ks = 144000/0.1
For those having NA litterature available, you will find the friction coefficient (Schoenherr + ATTC correction) roughly as: cf = 0.0023. This gives a friction force of Ff = 6438 N (compared to 6300 N). The stagnation force will be the same in both cases.

So, at 3 m/s, the total force is ~20000 N/pipe. Now that speed (6 knots) is probably too high. If we use 2 m/s, there is plenty of time for one round trip per hour, including mooring and load securing. Since all forces depend on speed^2, the total “fluid” force at 2 m/s will be 8740 N/pipe. To this add current, 1 m/s and you are back to 20 kN/pipe. Then add friction to rails (as noted by Gonzo) and safety margins for wind and weather. You would probably need maximum 30 kN/pipe.

The tug dimensioning factor is then how many pipes have to be delivered each trip. Say that 5 lengths/hour can be connected, then the tug must have a minimum towing force of 150 kN at 6 knots. The required transport power is then 150 * 3 kW.

Propeller efficiency during towing and trawling is low, due to low advance coefficient, so in spite of what TANSL's computer model states, use 40 á 45 % as propulsion efficiency. The towing power is then ~1100 kW for this example.

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### TANSLSenior Member

Interesting calculation, @baeckmo.
The 65% efficiency that I have considered has not been decided by my computer but by myself. If it is an error, it is my error. Fortunately those things still can not be deduced by a computer, they are in the hands of the designer.

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### Ali GolshaniJunior Member

Hi all, does above paragraph work for me?

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### Ali GolshaniJunior Member

No, it is Tugboat. Thanks

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### Ali GolshaniJunior Member

Thanks a lot.

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### gonzoSenior Member

I believe the maximum friction will be between the pipes and the steel guides.

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### baeckmoHydrodynamics

One issue that is not dealt with yet is the downforce created by the flow around the square (plugged) pipe end forward. Neither the "basic" model nor the Holtrop Mennen regression model gives any info on this (btw I think the H+M model is limited to far lower length/beam ratios than we have in this case). To satisfy my own curiosity, I made some rude CFD-simulations with semisubmerged single, three and five pipes; the latter two in flat bundles.

The result (at 3 m/s) shows good agreement with the "basic engineering" drag force for the single pipe; 19200 N. In addition there is a downforce of 6540 N, acting over the forward 3 to 4 meters. This phenomenon is well known among tug crews, and will limit the towing speed in order to avoid "digging in", or diving.

The drag for bundled three is 52020 N with a downforce of 36300 N, i.e. a reduction in drag/pipe, but a substantial increase in diving force. For five pipes in a flat bundle, the drag is 88500 N and the diving force has increased to 64400 N (!). This means that it is not practical to tow bundles of five at 3 m/s (6 knot), even if the tug can deliver the towing force. Again, this force is proportional to speed^2; so I would estimate 4 knot to be about the maximum towing speed for flat bundles.

That said, I have to keep nagging about propeller efficiency during towing. This value has a major influence over the operational conditions, safety and economy. It is not ok to pick a value from the end of the wet finger in the wind. During trawling operation at speeds around 3 to 4 knots, propeller efficiencies as low as 25 to 28 % are normal with fixed pitch propellers. Towing with CP prop in Kort nozzle and slightly higher speed may result in 40-ish percentages. Stating 65% efficiency at towing is misleading and unrealistic. Sometimes we have to "guesstimate" in engineering, but in this case we have accurate knowledge to fall back upon when it comes to propulsion.

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### Ad HocNaval Architect

Indeed

How true...and based upon "sound" engineering

Dont forget, there will be a small amount of residuary resistance to add too. At these low Fns, merely adding 10% to the total, should suffice.

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### Ali GolshaniJunior Member

Hi and thanks for the reply.
Due to existing balance concrete ballasts, the drag force between concrete ballast and water in towing operation should be calculated and added to the calculations. I attached the drawing of pipeline+ ballast section. The ballast are every 45m. Can we consider a modified drag coefficient for pipeline to consider the effect of ballasts indirectly ?
Thanks

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### Squidly-DiddlySenior Member

When he says "add them, not subtract" he means you might be "adding" vectors going in different directions, which most people would say is "subtracting".

Its sort of like accounting. You "add" both the positive deposits and negative withdraws together.

However, 10 tons pushing one way and 10 tons pushing the other way might add up to zero but its not the same as "no force involved", just like an account with a million dollar deposit and million dollar withdrawal ain't quite the same as "no activity".

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### TANSLSenior Member

Yes, forces are vectors and then "adding forces" means "adding vectors".
If you subtract a negative scalar value from a set of scalars, you are probably adding it. English semantics is not my strong point.
About banks account I'm afraid not to know much.

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### Ali GolshaniJunior Member

Hi,
I am trying to understand your calculation.
1. I think lamdba is 0.009 not 0.09. Would you plz confirm it?
2. For Fs, could you plz show me the detailed calculation? I can not reach to 13365 N.
3. Could you plz show me the detailed calculation of Schoenherr method?
4. ِDid you use Moody diagram for calculation of lamdba ?
Thanks

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### baeckmoHydrodynamics

Hello Ali,
On the lambda issue: there is a writing error in the text, you are correct, it should say 0,009! But the calculation is done with the correct value all along. I'll ask the moderator if it is possible to correct the text in post no 16. Thank you for noting!

I used a version of the Moody diagram, that frankly is a bit too small in scale, but in reality, the sum of uncertainties in the other factors involved will drown the finer details of friction calculation. I'll get back to you with the calc later.

Regarding the ballast blocks, you can probably calculate the drag as a drag factor (roughly ~1) times effective frontal area of each block times number of blocks on each pipe times the dynamic pressure. The blocks are longitudinally spaced so that they would not influence each other to a measurable degree.

But then I must ask how these blocks will behave when the pipe stack is moving along a row of poles or ducd'albes; how do you manage them not to get stuck?

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### Ali GolshaniJunior Member

Hi and thanks for reply.
These ballasts are fixed to the pipe and exist every 45m and their function is keeping the balance of pipes against rotation. They are different than the concrete ballasts used for sinking purpose. Therefore, they do not get stuck. Please see attached drawing of the pipeline cross-section.
You have considered Cd roughly 1. Would you please give us a reference as BS code says 2 for square cross section (see attached page).
Some more questions apart from my previous ones (I am waiting especially for your "submerged area" calculation for FS as well):
5. Is 0.1 mm appropriate for "surface roughness" for HDPE (High Density Polyethylene) pipelines?
6. Have you considered the force needed for start of the movement (apart from friction between trolleys and rails that you mentioned, there is a F=ma force)? Please look at our attached spreadsheet in this regards.
7. What do we need to consider as friction coefficient (static or dynamic) between trolleys and rails. Is it neglect-able? Please see attached photo.
8. You mentioned that there is Wave Making Resistance (WMR) which can be neglected for L=144m, what about L=300 and L=322? I think there are also negotiable as there is a direct relation between waterline length and WMR. Larger waterline means less WMR.
And last, for your info only one single pile is moved in our project not bundle ones.
Thanks for your help
Ali

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### baeckmoHydrodynamics

A quick answer "on the run": the Cd in your note refers to 2-dimensional objects.

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